Question

This is an ADDMATHS question!! It's a kinematics problem. Please help me for both a(i) and a(ii).

Answer 1

a(i). Since you are given a velocity v. time graph, the distance will be represented by:
$\vec{s}(t) = \int_{a}^{b}\left \| \vec{v}(t) \right \|$

In this case, however, we can just use simple geometry to evaluate the area under the graph v(t). I split it up into 2 trapezoids, and 1 rectangle. So, the area will be as follows:
$A_t = A_{t1}+A_{t2}+A_r$
$A_{t1} = \frac{30+15}{2}(10) = 225m$
$A_{t2} = \frac{15+25}{2}(15) = 300m$
$A_r = 25(30) = 750m$
$A_t = 750+300+225 = 1275m$

So, the particle traveled a total of 1275m assuming it never turned back (because it says to calculate distance).

a(iii). Deceleration is a word for negative acceleration. Acceleration is the first derivative of velocity, and so deceleration is too. So, we just need to find the slope of the line that passes through t = 30 because it has a linear slope (meaning the slope doesn't change). So, we can just use simple algebra instead of calculus to figure this out. Recall from algebra that slope (m):
$m = \frac{y_2-y_1}{x_2-x_1}$

So, let's just pick values. I'm going to pick (25, 30) and (35, 15). Let's plug and chug:
$m = \frac{15-30}{35-25} = -\frac{15}{10} = -\frac{3}{2}$

Since it's a negative value, this means that acceleration is negative but deceleration is positive (because deceleration is negative acceleration). So, your answer is: The deceleration of the particle at t = 30s is 3/2 or 1.5.